1D kinematics multi-system dog cat

Question

A dog is running toward a cat that is 1.5 m away. Initially at rest, the cat sees the dog running toward her and accelerates away from the dog at 0.85 m/s/s directly toward a small pet door that is another 1.5 m in front of the cat (too small for the dog, but OK for the cat to fit through). Seeing the small window ahead, the dog (initially moving at 2.1 m/s) begins to slow down at 0.2 m/s/s. We need to figure out if the cat makes it out of the door before the dog arrives, and so how far does the dog go by the time the cat reaches the door? Report your answer to the tenths decimal place.

Answer 1

So i can't draw an ideal picture, but because these 3 objects are standing in a line, this "text-pic" may work

petdoor — cat — dog

with each " —" = 1.5m. Let's say the petdoor resides at x=0m, the cat (initially) at x=1.5m, and the dog (initially) at x=3m

using x = x₀ + v₀t + a₀ t²/2 , where x₀ is the initial position of the object, v₀ the initial velocity, t is time given in seconds, and a₀ the initial (and in our case only) acceleration.

The cat's position is described by x_cat = 1.5m - 1/2(2.2m/s²)t²

it's initial velocity is 0, so the middle term disappears. It's also accelerating towards the door, which is negative (at least how we described it)

The dog's position is described by x_dog = 3m - (2.1m/s)t + 1/2(2.2m/s²)t²

its initially going towards the door, but then begins slowing down (aka the positive acceleration)

We know at x=0 (at the petdoor), the time for each to arrive is

x_cat = 0 = 1.5m - 1/2(.85m/s²)t²

--> t² = 3m/(.85m/s²) --> t_{cat reachs door} ≈ 1.878seconds (or a negative time, which doesn't make sense in this scenario), or 1.9 seconds to the nearest tenth

when calculating the dog's time of arrival at the petdoor, we note

x_dog = 0 = 3m - (2.1m/s)t + 1/2(.2m/s²)t² and using the quadratic equation we get

--> t_dog = [2.1 ± √(2.1² -4(.2)(3))] / (2*.2) = 1.5s or 19.5s

because we care about when the dog first reaches the petdoor(and the dog can't fit the thru it anyway), we care about the first time, 1.5s.

the dog indeed beats the cat to the petdoor, as it arrives there at a sooner time (poor cat)

Answer 2

I'm going to assume the dog instantaneously begins decelerating at the same time the cat starts acceleration and that all occurs at the distances given.

I always like to start with a picture:

The time it takes for the dog to get to the pet door can be calculated using the kinematic equation:

x = v_it + 1/2at²

Plug in the known values to get:

3 = 2.1t + 1/2(-0.2)t²

0.1t² - 2.1t + 3 = 0

Using the quadratic equation, solve for t:

t = [-(-2.1) ± √((-2.1)² - 4(0.1)(3))]/(2•0.1)

t = 1.542 seconds and t = 19.458 seconds. Considering these, it's clear to see that the second answer would be the time if the dog continued to slow to the point he stopped and reversed and then came back to the pet door so this is not the time we are interested in. So 1.542 seconds.

The time that the cat gets to the pet door would be calculated from the same equation:

1.5 = 0t + (1/2)(0.85)t²

0.425t² = 1.5

t² = 3.529

t = √3.529 = 1.879 seconds

Looks like the cat is toast.

Answer 3

Dog |------- 1.5m -------| Cat |------- 1.5m -------| Pet Door

|--------------------- 3.0m -----------------------| Origin

Assumptions:

Position x

Velocity v

Acceleration a

Let's put the origin of the graph at the pet door.

Anything to the left of the door is negative.

Givens:

Dog:

x_d = -3.0 m

v_d = 2.1 m/s

a_d = -0.2 m/s^2

Cat:

x_c = -1.5 m

v_c = 0 m/s (rest)

a_c = +0.85 m/s^2

Cat:

Let's focus on the time it takes the cat to reach the pet door.

The general equation is:

x(t) = x_0 + v_0 * t + (1/2) * a * t^2

where x_0 is x(0) or the initial position and v_0 is the initial velocity or v(0).

x_c(t) = x_c0 + v_c0 * t + (1/2) * a_c * t^2

x_c(t) = -1.5 + 0 * t + (1/2) * 0.85 * t^2

x_c(t) = -1.5 + 0.425 * t^2



x_c(t) = 0 when the cat reached the pet door.

0 = -1.5 + 0.425 * t^2

0.425 * t^2 = 1.5

t^2 = 1.5 / 0.425 ≈ 3.5294

t = SQRT(3.5294) ≈ ±1.8787 seconds

Therefore, the cat reaches the pet door at 1.8787 seconds.

Dog:

Now let's focus on the dog.

x_d(t) = x_d0 + v_d0 * t + (1/2) * a_d * t^2

x_d(t) = -3.0 + 2.1 * t + (1/2) * (-0.2) * t^2

x_d(t) = -3.0 + 2.1 * t - (0.1) * t^2

Where is the dog when the cat reaches the cat door?

x_d(1.8787) = -3.0 + 2.1 * (1.8787) - (0.1) * (1.8787)^2

x_d(1.8787) = -3.0 + 3.9453 - 0.3530

x_d(1.8787) = 0.5923 m

Since this is positive, the dog reached the pet door before the cat and is already half a meter past it.

Therefore the cat is toast.

X_c = X_d:

Setting the cat and dog equations equal to each other will tell us at what time and position the cat and dog met or intersected.

x_c(t) = x_d(t)

-1.5 + 0.425 * t^2 = -3.0 + 2.1 * t - (0.1) * t^2

Subtract everything on the right with the left to get:

1.5 - 2.1*t + (0.425) * t^2 + (0.1) * t^2 = 0

1.5 - 2.1*t + 0.525 * t^2 = 0

0.525 * t^2 - 2.1*t + 1.5 = 0

Using the quadratic equation so solve for t, we get

t = 0.9310 seconds and 3.0690 seconds

3 seconds is too late, so choose t = 0.9310 seconds.

Therefore, the dog reaches that cat after 0.9310 seconds or 0.9 seconds.

Cat's Position at Meeting:

x_c(t) = -1.5 + 0.425 * t^2

x_c(0.9310) = -1.5 + 0.425 * (0.9310)^2

x_c(0.9310) = -1.1316 m

Dog's Position at Meeting:

x_d(t) = -3.0 + 2.1 * t - (0.1) * t^2

x_d(0.9310) = -3.0 + 2.1 * (0.9310) - (0.1) * (0.9310)^2

x_d(0.9310) = -1.1316 m

Therefore the dog catches the cat at 1.1 meters away from the pet gate.